Download Sample problems Chap 21 Cutnell

Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
CHAPTER 21 Magnetism
CONCEPTUAL QUESTIONS
Cutnell & Johnson 7E
3. ssm A charged particle, passing through a certain region of space, has a velocity whose
magnitude and direction remain constant, (a) If it is known that the external magnetic field is
zero every where in this region, can you conclude that the external electric field is also zero?
Explain. (b) If it is known that the external electric field is zero everywhere, can you
conclude that the external magnetic field is also zero? Explain.
3.
SSM REASONING AND SOLUTION A charged particle, passing through a certain
region of space, has a velocity whose magnitude and direction remain constant.
a. If it is known that the external magnetic field is zero everywhere in the region, we can
conclude that the electric field is also zero. Any charged particle placed in an electric
field will experience a force given by F = qE, where q is the charge and E is the electric
field. If the magnitude and direction of the velocity of the particle are constant, then the
particle has zero acceleration. From Newton's second law, we know that the net force on
the particle is zero. But there is no magnetic field and, hence, no magnetic force.
Therefore, the net force is the electric force. Since the electric force is zero, the electric
field must be zero.
b. If it is known that the external electric field is zero everywhere, we cannot conclude
that the external magnetic field is also zero. In order for a moving charged particle to
experience a magnetic force when it is placed in a magnetic field, the velocity of the
moving charge must have a component that is perpendicular to the direction of the
magnetic field. If the moving charged particle enters the region such that its velocity is
parallel or antiparallel to the magnetic field, it will experience no magnetic force, even
though a magnetic field is present. In the absence of an external electric field, there is no
electric force either. Thus, there is no net force, and the velocity vector will not change
in any way.
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7. The drawing shows a top view of four interconnected chambers. A negative charge is fired
into chamber 1. By turning on separate magnetic fields in each chamber, the charge can be
made to exit from chamber 4, as shown. (a) Describe how the magnetic field in each chamber
should be directed. (b) If the speed of the charge is v when it enters chamber 1, what is the
speed of the charge when it exits chamber 4? Why?
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
7.
REASONING AND SOLUTION The drawing
shows a top view of four interconnected
chambers. A negative charge is fired into
chamber 1. By turning on separate magnetic
fields in each chamber, the charge is made to exit
from chamber 4.
4
3
F
F
F
F
a. In each chamber the path of the particle is one1
2
quarter of a circle. The drawing at the right also
shows the direction of the centripetal force that
v
must act on the particle in each chamber in order
–q
for the particle to traverse the path. The charged
particle can be made to move in a circular path by launching it into a region in which
there exists a magnetic field that is perpendicular to the velocity of the particle.
Using RHR-1, we see that if the palm of the right hand were facing in the direction of
F in chamber 1 so that the thumb points along the path of the particle, the fingers of the
right hand must point out of the page. This is the direction that the magnetic field must
have to make a positive charge move along the path shown in chamber 1. Since the
particle is negatively charged, the field must point opposite to that direction or into the
page. Similar reasoning using RHR-1, and remembering that the particle is negatively
charged, leads to the following conclusions: in region 2 the field must point out of the
page, in region 3 the field must point out of the page, and in region 4 the field must
point into the page.
b. If the speed of the particle is v when it enters chamber 1, it will emerge from chamber
4 with the same speed v. The magnetic force is always perpendicular to the velocity of
the particle; therefore, it cannot do work on the particle and cannot change the kinetic
energy of the particle, according to the work-energy theorem. Since the kinetic energy is
unchanged, the speed remains constant.
15. For each electromagnet at the left of the drawing, explain whether it will be attracted to
or repelled from the permanent magnet at the right.
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
15. REASONING AND SOLUTION The figure below shows the arrangements of electromagnets and magnets.
(a)
S
N
N
S
S
N
N
S
(b)
We can determine the polarity of the electromagnets by using RHR-2. Imagine holding
the current-carrying wire of the electromagnet in the right hand as the wire begins to coil
around the iron core. The thumb points in the direction of the current. For the
electromagnet in figure (a), the fingers of the right hand wrap around the wire on the left
end so that they point, inside the coil, toward the right end. Thus, the right end of the
coil must be a north pole. Similar reasoning can be used to identify the north and south
poles of the electromagnet in figure (b). The results are shown in the figure above.
Since the like poles of two different magnets repel each other and the dissimilar poles of
two different magnets attract each other, we can conclude that in both arrangements, the
electromagnet is repelled from the permanent magnet at the right.
CHAPTER 21 MAGNETIC FORCES
AND MAGNETIC FIELDS
Samples of solutions to Problems from chapter 21Cutnell
& Johnson 7E
4. When a charged particle moves at an angle of 25° with respect to a magnetic field, it
experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect
to this field will this particle, moving at the same speed, experience a magnetic force of
magnitude 2F?
4.
REASONING According to Equation 21.1, the magnetic force has a magnitude of
F = q vB sin θ, where q is the magnitude of the charge, B is the magnitude of the
magnetic field, v is the speed, and θ is the angle of the velocity with respect to the field.
As θ increases from 0° to 90°, the force increases. Therefore, the angle we seek must lie
between 25° and 90°.
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
SOLUTION Letting θ1 = 25° and θ2 be the desired angle, we jcan apply Equation 21.1
to both situations as follows:
F = q vB sin θ
1
and
Situation 1
2 F = q vB sin θ 2
Situation 2
Dividing the equation for situation 2 by the equation for situation 1 gives
2 F q vB sin θ 2
=
F
q vB sin θ1
or
sin θ 2 =2sin θ1 =2sin 25° = 0.85
θ 2 = sin −1 ( 0.85 ) = 58°
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11. A charged particle enters a uniform magnetic field and follows the circular path shown in
the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle’s
speed is 140 m/s, the magnitude of the magnetic field is 0.48 T, and the radius of the path is
960 m. Determine the mass of the particle, given that its charge has a magnitude of
.
11. REASONING
a. The drawing shows the velocity v of the particle at the top
of its path. The magnetic force F, which provides the
centripetal force, must be directed toward the center of the
circular path. Since the directions of v, F, and B are known,
we can use Right-Hand Rule No. 1 (RHR-1) to determine if
the charge is positive or negative.
B (out of paper)
v
F
b. The radius of the circular path followed by a charged
particle is given by Equation 21.2 as r = mv / q B . The mass
m of the particle can be obtained directly from this relation, since all other variables are
known.
SOLUTION
a. If the particle were positively charged, an application of RHR-1 would show that the
force would be directed straight up, opposite to that shown in the drawing. Thus, the
charge on the particle must be negative .
b. Solving Equation 21.2 for the mass of the particle gives
m=
q Br
8.2 × 10−4 C ) ( 0.48 T )( 960 m )
(
=
= 2.7 × 10−3 kg
140 m/s
v
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
19. ssm Review Conceptual Example 2 as an aid in understanding this
problem. The drawing shows a positively charged particle entering a 0.52-T
magnetic field (directed out of the paper). The particle has a speed of 270
m/s and moves perpendicular to the magnetic field. Just as the particle enters
the magnetic field, an electric field is turned on. What must be the
magnitude and direction of the electric field such that the net force on the
particle is twice the magnetic force?
19. SSM REASONING AND SOLUTION According to Right-Hand Rule No. 1, the
magnetic force on the positively charged particle is toward the bottom of the page in the
drawing in the text. If the presence of the electric field is to double the magnitude of the
net
force
on
the
charge,
the
electric
field
must
also
be
directed toward the bottom of the page . Note that this results in the electric field
being perpendicular to the magnetic field, even though the electric force and the
magnetic force are in the same direction.
Furthermore, if the magnitude of the net force on the particle is twice the magnetic force,
the electric force must be equal in magnitude to the magnetic force. In other words,
combining Equations 18.2 and 21.1, we find
q E = q vB sin θ , with
sin θ = sin 90.0° = 1.0 . Then, solving for E
E = vB sin θ = ( 270 m / s)(0.52 T)(1.0) = 140 V / m
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29. A square coil of wire containing a single turn is placed in a uniform
0.25-T magnetic field, as the drawing shows. Each side has a length of
0.32 m, and the current in the coil is 12 A. Determine the magnitude of
the magnetic force on each of the four sides.
29. REASONING AND SOLUTION The force on each side can be found from F = ILB sin
θ. For the top side, θ = 90.0°, so
F = (12 A)(0.32 m)(0.25 T) sin 90.0° = 0.96 N
The force on the bottom side (θ = 90.0°) is the same as that on the top side, F =
0.96 N .
For each of the other two sides θ = 0°, so that the force is F = 0 N .
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
41. ssm www The rectangular loop in the drawing consists of 75
turns and carries a current of I = 4.4 A. A 1.8-T magnetic field is
directed along the +y axis. The loop is free to rotate about the z
axis. (a) Determine the magnitude of the net torque exerted on
the loop and (b) state whether the 35° angle will increase or
decrease.
41. SSM WWW REASONING The torque on the loop is
given by Equation 21.4, τ = NIABsin φ . From the drawing
in the text, we see that the angle φ between the normal to the plane of the loop and the
90° − 35° = 55° .
The
area
of
the
loop
is
magnetic
field
is
2
0.70 m × 0.50 m = 0.35 m .
SOLUTION
a. The magnitude of the net torque exerted on the loop is
τ = NIAB sin φ = (75)(4.4 A)(0.35 m 2 )(1.8 T) sin 55° = 170 N ⋅ m
b. As discussed in the text, when a current-carrying loop is placed in a magnetic field,
the loop tends to rotate such that its normal becomes aligned with the magnetic field.
The normal to the loop makes an angle of 55° with respect to the magnetic field. Since
this angle decreases as the loop rotates, the 35° angle increases .
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59. The drawing shows an end-on view of three wires. They are long,
straight, and perpendicular to the plane of the paper. Their cross sections lie
at the corners of a square. The currents in wires 1 and 2 are I1 = I2 = I and
are directed into the paper. What is the direction of the current in wire 3, and
what is the ratio I3/I, such that the net magnetic field at the empty corner is
zero?
59. REASONING AND SOLUTION The currents in wires 1 and 2 produce the magnetic
fields B1 and B2 at the empty corner, as shown in the following drawing. The directions
of these fields can be obtained using RHR-2. Since there are equal currents in wires 1
and 2 and since these wires are each the same distance r from the empty corner, B1 and
B2 have equal magnitudes. Using Equation 21.5, we can write the field magnitude as
B1 = B2 = µ 0 I / 2 π r . Since the fields B1 and B2 are perpendicular, it follows from the
b g
Pythagorean theorem that they combine to produce a net magnetic field that has the
direction shown in the drawing at the right and has a magnitude B1+2 given by
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
B3
FG µ I IJ + FG µ I IJ
H 2π r K H 2π r K
2
B1+ 2 =
B12 + B22 =
0
0
2
=
Wire 1
X
2 µ0I
2π r
B2
r
B1
The current in wire 3 produces a field B3 at the empty
corner. Since B3 and B1+2 combine to give a zero net
Wire 3
field, B3 must have a direction opposite to that of B1+2.
Thus, B3 must point upward and to the left, and RHR-2
indicates that
B1+2
X
Wire 2
the current in wire 3 must be directed out of the plane of the paper .
Moreover, the magnitudes of B3 and B1+2 must be the same. Recognizing that wire 3 is a
distance of d =
B3 = B1+ 2
or
r2 +r2 =
µ0I3
2π
d
2r
i
2 r from the empty corner, we have
=
2 µ0I
2π r
so that
I3
I
=2
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60. The wire in Figure 21.40 carries a current of 12 A. Suppose that a second long, straight
wire is placed right next to this wire. The current in the second wire is 28 A. Use Ampère’s
law to find the magnitude of the magnetic field at a distance of r = 0.72 m from the wires
when the currents are (a) in the same direction and (b) in opposite directions.
60. REASONING Since the two wires are next to each other, the net magnetic field is
everywhere parallel to ∆A in Figure 21.40. Moreover, the net magnetic field B has the
same magnitude B at each point along the circular path, because each point is at the same
distance from the wires. Thus, in Ampère's law (Equation 21.8), B|| = B , I = I1 + I 2 ,
and we have
ΣB|| ∆A = B ( Σ∆A ) = µ0 ( I1 + I 2 )
But Σ∆A is just the circumference (2πr) of the circle, so Ampère's law becomes
B ( 2π r ) = µ0 ( I1 + I 2 )
This expression can be solved for B.
SOLUTION
a. When the currents are in the same direction, we find that
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Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E
(
)
–7
µ0 ( I1 + I 2 ) 4π × 10 T ⋅ m/A ( 28 A + 12 A )
B=
=
= 1.1 × 10 –5 T
2π r
2π ( 0.72 m )
b. When the currents have opposite directions, a similar calculation shows that
(
)
µ ( I – I ) 4π × 10 T ⋅ m/A ( 28 A–12 A )
B= 0 1 2 =
= 4.4 × 10 –6 T
2π r
2π ( 0.72 m )
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65. ssm A long solenoid has 1400 turns per meter of length, and it carries a current of 3.5 A.
A small circular coil of wire is placed in side the solenoid with the normal to the coil oriented
at an angle of 90.0° with respect to the axis of the solenoid. The coil consists of 50 turns, has
an area of 1.2 × 10–3 m2, and carries a current of 0.50 A. Find the torque exerted on the coil.
–7
65. SSM REASONING The coil carries a current and experiences a torque when it is
placed in an external magnetic field. Thus, when the coil is placed in the magnetic field
due to the solenoid, it will experience a torque given by Equation 21.4: τ = NIABsin φ ,
where N is the number of turns in the coil, A is the area of the coil, B is the magnetic
field inside the solenoid, and φ is the angle between the normal to the plane of the coil
and the magnetic field. The magnetic field in the solenoid can be found from Equation
21.7: B = µ0 nI , where n is the number of turns per unit length of the solenoid and I is
the current.
SOLUTION The magnetic field inside the solenoid is
B = µ0 nI = (4π × 10 –7 T ⋅ m/A) (1400 turns/m ) (3.5 A)=6.2 × 10 –3 T
The torque exerted on the coil is
τ = NIAB sin φ = (50)(0.50 A)(1.2 ×10 –3 m 2 )(6.2 ×10 –3 T)(sin 90.0°)= 1.9 × 10 –4 N ⋅ m
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